1. is divided into simple known graphs.

As shown in the figure, the area of the shaded part can be divided into two triangles instead of a formula.

So the shadow area =4*6/2+4*4/2=20.

2. Combination of rectangle and semicircle

Find the shadow area

Solution 1: First output the blank area in the lower right corner, that is, S 1 in the figure.

Multiplex △ABC area -s 1 machine shadow area

S 1= square area-sector area =2? -π2? /4=4-π

S△ABC=2*4/2=4

So the shadow area =4-(4-π)=π.

Solution 2: the diagonal line of long hair intersects the midline at point O.

Obviously, it can be seen that △COE and △AOF are completely equal, and the shadow part can be cut in the same area. As shown in the figure, the green part is moved to the red part and reassembled into a complete sector.

So the shadow area =π2? /4=π

3. Give me a question that seems difficult to start or confusing to calculate?

For a square with a side length of a, make a circle with each vertex as the center, make the circle pass through the center of the square, and find the shadow area.

(expressed by the expression of a)

I will show you the benefits of excavation and filling directly with graphics in this problem.

Cut the diagonal of the eye white square in the red part of the left picture into eight congruent parts, four of which are spliced to the red part of the right picture and the other four to the green part.

Obviously, the area of the shaded part in the above picture is twice that of the four white semicircles plus the red part (square-circle).

So this picture becomes again.

Now it's two circles on the right+red. Obviously, these two circles just fill the blank part of the square.

Divided into two square areas.

Finally, the complex shadow part is equivalent to the area of two squares, and it can be calculated by finding the side length of the square.

As can be seen from the picture, the side length of this square is the diagonal of the square with side length A given in the title.

It seems that I didn't learn to take the root sign in primary school, so you can calculate the square of the side length of a big square according to the square of the hypotenuse of a right triangle = the sum of the squares of the right sides.

What about the big diamond B? =2a？

The area of the big square =b? =2a？

So the shaded area required by the topic =2b? =4a？

Like the example above, isn't it more interesting to do? To do this kind of questions well, we must observe more, think more, consider more ways to answer them, find more questions to practice, and finally summarize them ourselves, so that it will be easy to encounter this kind of questions again in the future.

I wish you progress in your study! !